Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k + 3}{k^2 + 6k - 40} \div \dfrac{k + 7}{k^2 + 3k - 28} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{k + 3}{k^2 + 6k - 40} \times \dfrac{k^2 + 3k - 28}{k + 7} $ First factor out any common factors. $z = \dfrac{k + 3}{k^2 + 6k - 40} \times \dfrac{k^2 + 3k - 28}{k + 7} $ Then factor the quadratic expressions. $z = \dfrac {k + 3} {(k - 4)(k + 10)} \times \dfrac {(k - 4)(k + 7)} {k + 7} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {(k + 3) \times (k - 4)(k + 7) } { (k - 4)(k + 10) \times (k + 7)} $ $z = \dfrac {(k - 4)(k + 7)(k + 3)} {(k - 4)(k + 10)(k + 7)} $ Notice that $(k - 4)$ and $(k + 7)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {\cancel{(k - 4)}(k + 7)(k + 3)} {\cancel{(k - 4)}(k + 10)(k + 7)} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $z = \dfrac {\cancel{(k - 4)}\cancel{(k + 7)}(k + 3)} {\cancel{(k - 4)}(k + 10)\cancel{(k + 7)}} $ We are dividing by $k + 7$ , so $k + 7 \neq 0$ Therefore, $k \neq -7$ $z = \dfrac {k + 3} {k + 10} $ $ z = \dfrac{k + 3}{k + 10}; k \neq 4; k \neq -7 $